the complete question in the attached figure
we have
[(x²-x-6)/x²]=[(x-6)/2x]+[(2x+12)/x]
we know that
the LCD (the least common denominator) for the given equation is -----> 2x²
step 1
multiply each term by this factor
2x²*[(x²-x-6)/x²]=2x²*[(x-6)/2x]+2x²*[(2x+12)/x]
2*[(x²-x-6)]=x*[(x-6)]+2x*[(2x+12)]
[(2x²-2x-12)]=[(x²-6x)]+[(4x²+24x)]
[(2x²-2x-12)]-[(x²-6x)]-[(4x²+24x)]=0
2x²-2x-12-x²+6x-4x²-24x=0
2x²-x²-4x²-2x+6x-24x-12=0
-3x²-20x-12=0
multiply by -1 the entire equation
3x²+20x+12=0
the answer is
3x²+20x+12=0
